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I am trying to prove the following:

If $F_1$ and $F_2$ are free modules on the same set $A$, there is a unique isomorphism between $F_1$ and $F_2$ which is the identity map on $A$.

The definition I am using is

An $R$-module $F$ is said to be free on a subset $A$ of $F$, if for every nonzero element $x\in F$, there exist unique nonzero elements $r_1,\cdots,r_n \in R$ and unique $a_1,a_2,\cdots,a_n \in A$ such that $x = r_1 a_1 + \cdots + r_n a_n$ for some $n \in \mathbb{Z}^+$.

Then regarding the universal mapping property:

For any set $A$, there is a free $R$-module $F(A)$ satisfying the following universal property:...

I actually tried the following:

Since $F_1$ is a free module on set $A$, then for every $x\in F_1$, it can be written as $x = r_1 a_1 + \cdots + r_n a_n$ for unique nonzero $r_i\in R$ and unique $a_i \in A$. Similarly, for every $y\in F_2$, $y = s_1 a_1 + \cdots + s_m a_m$.

I wanted to say that let's just consider

$$\phi: r_1 a_1 + \cdots + r_n a_n \longrightarrow r_1 a_1 + \cdots + r_n a_n$$

But then I am not sure if $r_1 a_1 + \cdots + r_n a_n$ is in $F_2$ or not? $F_2$ being a free module does not imply that all possible combinations $\sum r_i a_i$ is in it, but only implies that if something is in it, it can be uniquely written in that form, right? How should I show surjectivity?

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    $\begingroup$ It would help to give the definition of "free module on a set" you are using, since there are many different possible meanings. For example, I would probably not define a free module on $A$ to necessarily have $A$ as a subset, but you seem to have assumed this. $\endgroup$ – Slade Nov 15 '16 at 17:58
  • $\begingroup$ To answer the question about being in $F_2$: if you are assuming that $A\subset F_2$, then all $R$-linear combinations of elements of $A$ are also in $F_2$, since $F_2$ is an $R$-module. $\endgroup$ – Slade Nov 15 '16 at 18:00
  • $\begingroup$ @Slade Edited. Basically I always see that there is a homomorphism between $F_1$ and $F_2$, but somehow I need to use universal property (perhaps) to show that it's actually an isomorphism $\endgroup$ – 3x89g2 Nov 15 '16 at 18:05
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    $\begingroup$ @AlexM No, according to his definition a free module on $A$ must contain $A$, which the examples you're giving are not guaranteed to. $\endgroup$ – Slade Nov 15 '16 at 18:21
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    $\begingroup$ @AlexM. It literally says that $A$ is a subset of $F$. $\endgroup$ – Slade Nov 15 '16 at 18:23
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$\renewcommand{\phi}{\varphi}$ Addendum

OP has given the definition of free modue he assumes. Let us prove that it satisfies the universal property I use below.

Let $f : A \to M$ a map, where $M$ is a module. Since every element of $F(A)$ can be uniquely written as $\sum_{i=1}^{n} r_{i} a_{i}$, for $r_{i} \in R$ and $a_{i} \in A$, a morphism $\phi: F(A) \to M$ such that $\phi(a) = f(a)$ for $a \in A$ (provided it exists) is well defined and uniquely defined as $$ \phi\left(\sum_{i=1}^{n} r_{i} a_{i}\right) = \sum_{i=1}^{n} r_{i} \phi(a_{i}). $$ Now it is not difficult to verify that this map is indeed a morphism $F(A) \to M$, $$ \phi\left(\sum_{i=1}^{n} r_{i} a_{i} + \sum_{i=1}^{n} s_{i} a_{i}\right) = \phi\left(\sum_{i=1}^{n} (r_{i}+s_{i}) a_{i}\right) = \sum_{i=1}^{n} (r_{i}+s_{i}) \phi(a_{i}) =\\= \sum_{i=1}^{n} r_{i} \phi(a_{i}) + \sum_{i=1}^{n} s_{i} \phi(a_{i}) = \phi\left(\sum_{i=1}^{n} r_{i} a_{i}\right) + \phi\left(\sum_{i=1}^{n} s_{i} a_{i}\right). $$


You should have defined a free module over $A$ as a module $F$ containing $A$ such that if $M$ is any module, and $f: A \to M$ is a map, then there is a unique module morphism $\phi : F \to M$ such that $\phi(a) = f(a)$ for each $a \in A$.

So if $F_{1}, F_{2}$ are two free modules over $A$, consider $f_{2} : A \to F_{2}$ to be the identity (or inclusion) map, i.e. $f_{2}(a) = a$ for $a \in A$. Since $F_{1}$ is free on $A$, there is a unique morphism $\phi_{2}: F_{1} \to F_{2}$ such that $\phi_{2}(a) = a$ for $a \in A$. Similarly, there is a unique morphism $\phi_{1}: F_{2} \to F_{1}$ such that $\phi_{1}(a) = a$ for $a \in A$.

Now the composition $\phi = \phi_{1} \circ \phi_{2}$ is a morphism $F_{1} \to F_{1}$ such that $\phi(a) = a$ for $a \in A$. Since $F_{1}$ is free, this is unique, and thus is the identity, as the identity also maps all elements $a \in A$ to $a$. Similary, $\phi_{2} \circ \phi_{1}$ is the identity on $F_{2}$, and thus $\phi_{1}, \phi_{2}$ are isomorphisms, one the inverse of the other.

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  • $\begingroup$ My book actually first defined free module in a different way, then showed that free module satisfies the universal property that you mentioned. Need some times digesting the last part. Thanks! $\endgroup$ – 3x89g2 Nov 15 '16 at 18:18
  • $\begingroup$ So in general, if composite of two homomorphisms is identity, then each of those is isomorphism? $\endgroup$ – 3x89g2 Nov 15 '16 at 18:30
  • $\begingroup$ $\renewcommand{\phi}{\varphi}$If both $\phi_{1} \circ \phi_{2}$ and $\phi_{2} \circ \phi_{1}$ are the identity on their respective domains, then they are bijective maps (a general fact in set theory), and thus isomorphisms. $\endgroup$ – Andreas Caranti Nov 15 '16 at 18:33
  • $\begingroup$ OK, now I realize why it's good to use universal property as definition. I am having trouble justifying "Since $F_1$ is free on $A_1$, there is a unique..." part because according to my definition, we do not know if that's true or not. We only know there is some $F(A)$ satisfying the property, but that might not be $F_1$... $\endgroup$ – 3x89g2 Nov 15 '16 at 19:00
  • $\begingroup$ You want to prove uniqueness of the free module up to isomorphism. So you take two free modules, and show they are isomorphic, that's it. $\endgroup$ – Andreas Caranti Nov 15 '16 at 19:26
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All linear combination $\sum_{i=1}^n r_ia_i$ are in $F_2$ because is a module and as such closed under addition and multiplication by ring elements and $F_2$ contains $A$.

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  • $\begingroup$ That's not obviously to me. Let $A = \{ a_1, a_2\}$. Let $R = \{r_1,r_2\}$. What if I consider $\{ r_1 a_1 ,r_2 a_1\}$? $\endgroup$ – 3x89g2 Nov 15 '16 at 18:22
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You have given the right definition of $\phi$. Let's be clear about the notation: Define $\phi(r_1\cdot a_1 + \cdots + r_n a_n) = r_1\odot a_1 \oplus \cdots \oplus r_n \odot a_n$, where I'm using $\cdot$ and $+$ for multiplication and addition in $F_1$, and $\odot$ and $\oplus$ for the multiplication in $F_2$. After all, they may be different.

You ask whether the right-hand side is "in $F_2$". Yes, certainly it is; it's an $R$-linear combination of elements of $F_2$, and $F_2$ is an $R$-module.

Note that $\phi$ is well-defined precisely because any element of $F_1$ can be written uniquely in the form $r_1\cdot a_1 + \cdots + r_n a_n$.

We should check that $\phi$ is really an $R$-module homomorphism. This is not completely trivial, but it should be intuitive: if $x=\sum_i r_i a_i$ and $y=\sum_i s_i a_i$, then $x+y=\sum_i (r_i + s_i) a_i$ and $r\cdot x = \sum_i (r r_i)\cdot a_i$. There is a little more work to be done here, involving the operations $\odot$ and $\oplus$, but I leave it to you.

Finally, why is $\phi$ a bijection? Well, this comes down to the fact that $F_2$ is a free module over $A$. In fact, constructing $\phi$ only required that $F_2$ is some $R$-module containing $A$, but to prove injectivity, we need to know that representations in $F_2$ are unique, and to prove surjectivity, we need to know that these representations exist for all elements of $F_2$.

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